Sunday, March 6, 2022

Recent Questions - Mathematics Stack Exchange

Recent Questions - Mathematics Stack Exchange

Volume of tetrahedron / dodecahedron in $m$ dimension

Posted: 06 Mar 2022 07:45 AM PST

What is the formula for calculating volume of tetrahedron in $m$ dimension ?
I am looking for equation dependent on radius of described circle.

I know equations only for $\text{2D}$ and $\text{3D}$ cases.

Tetrahedron:

For $\text{2D}$ it is: $$ V(r) = \frac{3 \sqrt{3}}{4} * r^2 $$

For $\text{3D}$ it is:
$$ V(r) = \frac{8 \sqrt{3}}{27} * r^3 $$

Dodecahedron :

For $\text{2D}$ it is: $$ V(r) = \frac{5}{4} \sqrt{ \frac{1}{2} (5 + \sqrt{5}) } * r^2 $$

For $\text{3D}$ it is:
$$ V(r) = \frac{2}{9} \sqrt{3} (5 + \sqrt{5}) * r^3 $$

I was looking fo the fromula for at least few hours without any effects.
Only what I can find is some info about count of verticles, etc.

The isomorphism of $ SO_4(\mathbb{R}) $ with $ SU_2 \otimes SU_2 $

Posted: 06 Mar 2022 07:42 AM PST

Let $ A,B $ be matrix groups (with entries in the same field). Then the tensor/Kronecker product $ A \otimes B $ is a matrix group and $$ \pi: A \times B \to A \otimes B $$ is a group homomorphism. Taking $ A=B=SU_2 $ we have a map $ \pi: SU_2 \times SU_2 \to SU_4 $ given by $$ (A,B) \mapsto A \otimes B $$ The only nontrivial element of the kernel is $ (-1,-1) $. So the image $ SU_2 \otimes SU_2 $ of $ \pi $ is a subgroup of $ SU_4 $ isomorphic to $ SO_4(\mathbb{R}) $.

Is $ SU_2 \otimes SU_2 $ conjugate to $ SO_4(\mathbb{R}) $ in $ SU_4 $?

Since matrices in $ SU_2 $ have real trace then all matrices in $ SU_2 \otimes SU_2 $ have real trace. So it is at least plausible that $ SU_2 \otimes SU_2 $ and $ SO_4(\mathbb{R}) $ are conjugate.

What are the steps to calculate this? (Fundamental theorem of calculus)

Posted: 06 Mar 2022 07:35 AM PST

"If $F(x)=\cos(-2x-4)\,\cdot\,2^{(10-x)/5}\, $ and $f(x)=F'(x)\, $ what is the value of $\int_{-4}^{2}f(x)\,dx\,. $"

The answer was given as $4.1$ but how did they get to this? I calculated the derivative of the $F(x)$ and got $2^{2x+1}(sin(-2x-4)+log(2)cos(-2x-4))$ and then when I plugged that into $\int_{-4}^{2}f(x)\,dx\, $ and ended up getting $-2.325$

How did they calculate the answer to be $4.1$?

Inequality involving expectation and L^p norm

Posted: 06 Mar 2022 07:48 AM PST

We denote by $\left\|. \right\|_{p}$ the usual $L^p$ norm.

Let :

  • $K$ a real valued function and $K_h:x\mapsto \frac{1}{h}K\left ( \frac{x}{h} \right )$ for $h> 0$.
  • $t$ a real valed function such that $\left\| t \right\|_{2} \leq 1$.
  • $X$ a real valued random variable with bounded density $f$ with respect to the Lebesgue measure.

According to this article (p.13) we have, for $h'\leq h$ : $$ \mathbb{E}\left [ \left ( \int \left ( K_{h'}-K_{h} \right )\left ( x-X \right )t\left ( x \right )\mathrm{d}x \right )^2 \right ] \leq \mathbb{E}\left [ \left ( \int |K_{h'}-K_{h}|\left ( x-X \right )\mathrm{d}x \right ) \right ] \mathbb{E}\left [ \left ( \int |K_{h'}-K_{h}|\left ( x-X \right )t^2\left ( x \right )\mathrm{d}x \right ) \right ] \leq \left\| K_{h'}-K_{h} \right\|_{1}^2\left\| f \right\|_{\infty}\left\| t \right\|_{2}^2 $$

I do not understand how to obtain that inequality. I tried to apply Cauchy-Schwartz's inequality $\left\| gh \right\|_{1} \leq \left\| g\right\|_{2}\left\| h \right\|_{2}$ for $g(x)=\sqrt{|K_{h'}-K_{h}|\left ( x-X \right )}$ and $h(x)=g(x)|t|(x)$ but everything still remains inside a singular expectation when I do that. I also tried to apply Cauchy-Schwartz with the scalar product $ \left< u\left ( X \right ),v\left ( X \right ) \right>= \mathbb{E}\left [ \int u\left ( X \right )v\left ( X \right )\right ]$ but then the square is outside of the expectation in the left hand side of the inequality.

The second inequality is also a bit unclear to me, I don't understand how to bound the second expectation.

Extremal of the functional J[u]

Posted: 06 Mar 2022 07:31 AM PST

Let $J[u]=\int_0^1({y''}^2-16x^2u')\text{dx}-u^2(1), \quad G=\{u'(1)=2,\; 2u'(0)+u(0)=1\}$

$\delta J[u](h)=\int_0^1(2u^{\text{IV}}+32x)h\text{dx}+2u''h'|_1-2u''h'|_0+(2u'''+16x^2)h|_0-(2u'''+16x^2+2u)h|_1$

$u^{\text{IV}}+16x=0$

$u_0=C_1+C_2x+C_3x^2+C_4x^3-\frac{2}{15}x^5$

How can I find constants $C_1, C_2, C_3, C_4$ satisfying given area G?

An uncountable product of $\mathbb{R}$ with itself is not metrizable (product topology).

Posted: 06 Mar 2022 07:43 AM PST

An uncountable product of $\mathbb{R}$ with itself isn't metrizable in the product topology

I wrote what I believe is an incorrect proof of this statement since I feel like I ignored the fact that the product is uncountable. However I can't find where is my mistake.

Cannot this proof "work" for countably infinite product of $\mathbb{R}$ ? (But aren't these products of $\mathbb{R}$ metrizable with the product topology ?).

Proof:

Let $J$ be an uncountable index set.

Let $(\mathbb{R}^J,\tau)$ be a topological space under the product topology.

Let $\mathcal{N}(x)$ denote the set of every neighbourhood of $x$ in $\mathbb{R}^J$.

Let $A\subset\mathbb{R}^J$ such that $\forall x \in A$, $x_\alpha = 1\text{ except for finitely many }\alpha\in J$,

We want tot show that $0\in\overline{A}$ ?

Let $U\in\mathcal{N}(0)\subset\mathbb{R}$. \begin{gather} U\in\tau \Rightarrow U=\prod_{\alpha\in J}{U_\alpha}\text{ with finitely many } U_\alpha \neq \mathbb{R} \end{gather} Let $\Gamma = \left\lbrace \beta\in J \vert U_\beta \neq \mathbb{R} \right\rbrace$

Then for $\beta\in\Gamma$ we know $U_\beta$ is open in $\mathbb{R}$ and $0\in U_\beta$.

So, \begin{gather} U_\beta = ]a_\beta; b_\beta[\text{ with }a_\beta < 0 < b_\beta \end{gather}

Let $y\in\mathbb{R}^J$ s.t. for $\alpha\in J$, \begin{gather} y_\alpha = \frac{b_\alpha - a_\alpha}{2}\text{, }\alpha\in\Gamma\\ y_\alpha = 1\text{, }\alpha\notin\Gamma \end{gather} Then $y\in A$ therefore $\exists y\in A$ s.t. $y\in U \cap A$.

Since $U\in\mathcal{N}(0)$ we have $0\in A' \subset \overline{A}$.

Now we must show that all sequences $(x_n)$ of $A$ do not converge to $0$.

Suppose that there is a sequence $(x_n)$ of $A$ s.t. $x_n \rightarrow 0$ \begin{gather} \forall U\in\mathcal{N}(0):\exists N\in\mathbb{N}:n>N\Rightarrow x_n\in U \end{gather} We note $x_{n,\alpha}$ the $\alpha$th coordinate of $x_n$.

Let $D_n=\lbrace i\in J \vert x_{n,i}\neq 1\rbrace$ (obviously $D_n$ is finite $\forall n\in\mathbb{N}$).

Then, let \begin{gather} \Gamma_n = \prod_{\alpha\in J}U_\alpha \end{gather} such that \begin{gather} U_\alpha = ]-|x_{n,\alpha}|;|x_{n,\alpha}|[\text{, }\alpha\in D_n\\ U_\alpha = \mathbb{R}\text{, }\alpha\notin\ D_n \end{gather} Then $\forall n\in\mathbb{N}$ we have $\Gamma_n\in\mathcal{N}(0)$ and $x_n\notin\Gamma_n$.

So $x_n \nrightarrow 0$, which further imply that every sequence $(x_n)$ of $A$ cannot converge to $0$.

Therefore $\mathbb{R}^J$ isn't metrizable.

How to solve for x = (2a/b) = b/(a/2)

Posted: 06 Mar 2022 07:29 AM PST

x = 2a/b = b/(a/2)

I am pretty lost in regards to how to approach this problem. If x is equal to 2a/b and 2a/b is equal to b/(a/2) then do I need to somehow get b/(a/2) into the form of 2a/b in order to solve?

Are there infinitely many square numbers with increasing digits? [duplicate]

Posted: 06 Mar 2022 07:30 AM PST

This is a question that came up while joking around with my friends, but now I am really intrigued by this question.

For sake of brevity, let's call square numbers with monotone increasing digits peculiar squares. Some examples of peculiar squares are $13^2 = 169$ (since $1 \leq 6 \leq 9$) and $15^2 = 225$. Question is, are there infintely many peculiar squares?

To tackle this question, I came up with a more generalized conjecture:

Peculiar Square Conjecture. For all $n \in \mathbb{N}$, there exists only finitely many peculiar squares in base $n$.

I first tried solving for $n=2$. This was pretty easy, since it is equivalent to proving that there are only finitely many squares of form $11\cdots 1_{(2)}$.

Then I tried solving for $n=3$. Simple number theory shows that $11\cdots 122 \cdots 2_{(3)}$ cannot be a square number. Thus, we only need to show that there are finitely many squares of form $11\cdots 1_{(3)}$. This was much easier said than done, and in the end I had to borrow the power of StackExchange. (Integer solutions of $3^n-1=2m^2$)

So up to this point, I know that the Peculiar Square Conjecture holds for $n = 2$ and $n = 3$, but I don't have clear idea of how to prove it for $n = 4$ or beyond. Any help or ideas would be much appreciated.

If an event with probability p occurs 2 out of 3 times, how can we find the probability that p > 0.5?

Posted: 06 Mar 2022 07:21 AM PST

If an event with probability "p" occurs 2 out of 3 times, how can we find the probability that p > 0.5?

Necessary condition for Open map in Inverse function theorem

Posted: 06 Mar 2022 07:34 AM PST

$f:\mathbb{R}^3 \to \mathbb{R}^3$ , $$f(x_1,x_2,x_3)=(e^{x_2cosx_1},e^{x_2sinx_1},2x_1-cosx_3)$$ and $E=(x_1,x_2,x_3)$ such that there exist an open subset U around $$(x_1,x_2,x_3)$$ such that f restricted to U is an open map. Then E is?

My approach: Function is continuously differentiable so I have used inverse function theorem and find the point in which Jacobian is non zero. The point were $R^3/(x_1,x_2,nπ)$. So at the point there will exist a neighborhood such that, f will be an open map at that neighborhood. Is there any other point namely on $(x_1,x_2,nπ)$. Is there any result guarantee such that Jacobian is zero at a point p and it will/will not have a neighborhood for which it's an open map? Suggest me some book for learning this topic, Inverse function theorem and thanks in advance.

Minimum Length of Circumradius $R$

Posted: 06 Mar 2022 07:43 AM PST

With usual notation, if in a triangle $\Delta ABC$, $a = 3$, $b = 4$ and circumradius R of is minimum, then the value of $[2rR]$ is (where [.] denotes greatest integer function)

My approach is as follow

$R = \frac{{abc}}{{4\Delta }} \Rightarrow R = \frac{{abc}}{{4rs}} \Rightarrow 2Rr = \frac{{abc}}{{2s}}$

$2s = a + b + c \Rightarrow 2s - 7 = c$

$ \Rightarrow 2Rr = \frac{{12c}}{{c + 7}} \Rightarrow 2Rr = \frac{{12}}{{1 + \frac{7}{c}}}$

Not able to proceed

find functors in the Rel category

Posted: 06 Mar 2022 07:34 AM PST

I'm learning Category Theory by Steve Awodey's book. in first exercise we have :

  1. The objects of Rel are sets, and an arrow $A \rightarrow B$ is a relation from $A$ to $B$, that is, a subset $R \subseteq A \times B .$ The equality relation $\{\langle a, a\rangle \in A \times A \mid a \in A\}$ is the identity arrow on a set $A$. Composition in Rel is to be given by $$ S \circ R=\{\langle a, c\rangle \in A \times C \mid \exists b(\langle a, b\rangle \in R \&\langle b, c\rangle \in S)\} $$ for $R \subseteq A \times B$ and $S \subseteq B \times C$.

(a) Show that Rel is a category.

(b) Show also that there is a functor $G$ : Sets $\rightarrow$ Rel taking objects to themselves and each function $f: A \rightarrow B$ to its graph, $$ G(f)=\{\langle a, f(a)\rangle \in A \times B \mid a \in A\} . $$

(c) Finally, show that there is a functor $C:$ Rel $^{\circ p} \rightarrow$ Rel taking each relation $R \subseteq A \times B$ to its converse $R^{c} \subseteq B \times A$, where, $$ \langle a, b\rangle \in R^{c} \Leftrightarrow\langle b, a\rangle \in R . $$

solution :

(a) Identity arrows behave correctly, for if $f \subset A \times B$, then $$ \begin{aligned} f \circ 1_{A} &=\left\{\langle a, b\rangle \mid \exists a^{\prime} \in A:\left\langle a, a^{\prime}\right\rangle \in 1_{A} \wedge\left\langle a^{\prime}, b\right\rangle \in f\right\} \\ &=\left\{\langle a, b\rangle \mid \exists a^{\prime} \in A: a=a^{\prime} \wedge\left\langle a^{\prime}, b\right\rangle \in f\right\} \\ &=\{\langle a, b\rangle \mid\langle a, b\rangle \in f\}=f \end{aligned} $$ and symmetrically $1_{B} \circ f=f$. Composition is associative; if $f \subseteq A \times B$, $g \subseteq B \times C$, and $h \subseteq C \times D$, then $(h \circ g) \circ f=\{\langle a, d\rangle \mid \exists b:\langle a, b\rangle \in f \wedge\langle b, d\rangle \in h \circ g\}$ $$ \begin{aligned} &=\{\langle a, d\rangle \mid \exists b:\langle a, b\rangle \in f \wedge\langle b, d\rangle \in\{\langle b, d\rangle \mid \exists c:\langle b, c\rangle \in g \wedge\langle c, d\rangle \in h\}\} \\ &=\{\langle a, d\rangle \mid \exists b:\langle a, b\rangle \in f \wedge \exists c:\langle b, c\rangle \in g \wedge\langle c, d\rangle \in h\} \\ &=\{\langle a, d\rangle \mid \exists b \exists c:\langle a, b\rangle \in f \wedge\langle b, c\rangle \in g \wedge\langle c, d\rangle \in h\} \\ &=\{\langle a, d\rangle \mid \exists c:(\exists b:\langle a, b\rangle \in f \wedge\langle b, c\rangle \in g) \wedge\langle c, d\rangle \in h\} \\ &=\{\langle a, d\rangle \mid \exists c:\langle a, b\rangle \in g \circ f \wedge\langle c, d\rangle \in h\} \\ &=h \circ(g \circ f) . \end{aligned} $$

for "b" we have :

  1. A functor from Set to Rel is a function mapping sets and functions (in Set) to sets and relations (in Rel).

  2. The map which sends every function $f$ to its graph $\Gamma(f)$ does send each function (in Set) to some relation (in Rel). All we need to prove is that it is "functorial"; that is, that it preserves identities and composition.

  3. And indeed, first we note that it behaves properly on objects: it sends functions $A\rightarrow B$ to relations on $A \times B$.

  4. And it preserves identities because it sends each identity function in Set $1_A : A \rightarrow A$ to the graph $\Gamma(1_A)$ in Rel; this graph is the identity relation $id_A : A\times A = \{\langle a,a\rangle : a\in A\}$.

  5. And it preserves composition: if $f:B\leftarrow A$ and $g:C\leftarrow B$, then $(g\circ f):C\leftarrow A$, and the graph of $(g\circ f)$ is actually the composition of the graph of $g$ and the graph of $f$. (The composition of two relations is defined in the expected way.)

  6. This shows that the graph construction is a functor.

how we can solve "c" ?

Showing that the operator $T$ is a bounded linear operator mapping $L^1[0, 1]$ to $c_0$

Posted: 06 Mar 2022 07:49 AM PST

For $f \in L^1[0,1]$ let $x_n = \int_0^1 f(t) t^n \: dt$, and let $T(f) = \{ x_n \}$. I want to show that the operator $T$ is a bounded linear operator mapping $L^1[0, 1]$ to $c_0$ (the space of real convergent sequences that converge to $0$) and determine the norm of $T$.

I have tried to go about this in a direct manner but haven't had any luck. For a bounded linear operator I am using the standard definition:

Let $X$ and $Y$ be normed linear spaces. A linear operator $T: X \to Y$ is bounded if there exists an $M \geq 0$ such that $||T u|| \leq M ||u||$ for all $u \in X$.

Lastly it is worth noting that I am considering the standard norm of $c_0$. That is to say $$||(x_1, x_2, x_3, \cdots)|| = \sup \{ |x_n| : n \in \mathbb{N} \}.$$

Solve $\sin(x)\cos(x)=\sin(x)+\cos(x)$

Posted: 06 Mar 2022 07:23 AM PST

My initial idea was $$(\sin(x)\cos(x))^2=1+2\sin(x)\cos(x)$$ Let $t=\sin(x)\cos(x)$; $$t^2=1+2t \quad\Leftrightarrow\quad t=1-\sqrt2$$ (since $1+\sqrt2>1$). I.e. $$\sin(x)\cos(x)=1-\sqrt2 \quad\Leftrightarrow\quad \tfrac12\sin(2x)=1-\sqrt2 \quad\Leftrightarrow\quad x=\tfrac{1}{2}\arcsin(2(1-\sqrt2))$$ but I didn't get any 'elegant' final solution. Any better ideas?

Why is the third derivative of cumulant generating function = skewness?

Posted: 06 Mar 2022 07:25 AM PST

From what I know, for random variable $X$, skewness is defined as

$$\mathbb{E}\left(\frac{X-\mathbb{E}(X)}{\sigma}\right)^3$$

or

$$\frac{\mathbb{\mathbb{E}}(X^3)-3\mathbb{E}(X)\sigma-\sigma^3}{\sigma^3}$$

where $\sigma$ is the standard deviation of $X$.

and the third derivative of cumulant generating function is

$$\mathbb{E}(X^3)-3\mathbb{E}(X)\mathbb{E}(X^2)+2[\mathbb{E}(X)]^3$$

These $2$ formulae look totally different, but why the third derivative of cumulant generating function is defined as the skewness of $X$?

Statement on Lie Group and Lie Algebra Homomorphisms

Posted: 06 Mar 2022 07:32 AM PST

I am reading J.J.Duistermaat and J.A.C.Kolk's Lie Groups. I cannot figure out (1.8.6). The textbook said and I quote:

Because $ad$ is a Lie algebra homomorphism: $\mathfrak g\to\mathfrak{gl(g)}$, one has $e^{ad(ad\,X)}\circ ad\,Y=ad(e^{ad\,X}\,Y)$.

I know that a Lie algebra homomorphism is a function that preserve the Lie bracket, but I cannot figure out how to prove the statement (1.8.6).

Is my LU Factorization Incorrect?

Posted: 06 Mar 2022 07:25 AM PST

I have the following matrix I am trying to decompose into it's respective $L$ and $U$ parts for $A = LU$.

So I have $$\begin{bmatrix} 1 && 4 && 3 \\ 0 && -10 && -5 \\ 0 && -8 && -4 \end{bmatrix}$$

$$ \left[ \begin{array}{cccccc|cccccc} 1 && 4 && 3 && && 1 && 0 && 1\\ 0 && -10 && -5 && && 0 && 1 && 0\\ 0 && -8 && -4 && && 0 && 0 && 1 \end{array} \right] $$

I notice that column $1$ is fine, and I need to make $R_3 C_2 = 0$ to achieve the upper triangle matrix on the left and the lower triangle matrix on the right.

So I do: $R_3 \longrightarrow -\frac{5}{4} R_3 + R_2$

$$ \left[ \begin{array}{cccccc|cccccc} 1 && 4 && 3 && && 1 && 0 && 1\\ 0 && -10 && -5 && && 0 && 1 && 0\\ 0 && 0 && 0 && && 0 && 1 && -5/4 \end{array} \right] $$

However, the answer key is telling me the answer is:

$$ \left[ \begin{array}{cccccc|cccccc} 1 && 4 && 3 && && 1 && 0 && 1\\ 0 && -10 && -5 && && 0 && 1 && 0\\ 0 && -8 && -4 && && 0 && 4/5 && 1 \end{array} \right] $$

Calculate the inverse Laplace transform by convolution

Posted: 06 Mar 2022 07:25 AM PST

Consider the following problem: determine $\displaystyle\mathscr{L}^{-1}\bigg[\frac{s^2+1}{s^2(s^2-4s+9)}\bigg]$ using formulas and using convolutions.

Using the formulas I found that the solution would be: $\displaystyle\frac{1}{81}\cdot\Big[9t+16\sqrt{5}e^{2t}sin(\sqrt{5}t)-4e^{2t}cos(\sqrt{5}t)+4\Big].$

But using convolutions we encountered problems. I thus considered: $H(s)=\displaystyle\frac{s^2+1}{s^2(s^2-4s+9)}, F(s)=\displaystyle\frac{s^2+1}{s^2}$ and $G(s)=\displaystyle\frac{1}{s^2-4s+9}\Rightarrow f(t)=\delta_t+t$ and $g(t)=\displaystyle\frac{1}{\sqrt{5}}e^{2t}sin(\sqrt{5}t)$ so $\displaystyle h(t)=f(t)*g(t)=\int_{0}^tf(t-\tau)g(\tau)d\tau=\int_{0}^{t}\Big(\delta(t-\tau)+t-\tau\Big)\cdot\Big(\frac{1}{\sqrt{5}}e^{2\tau}sin(\sqrt{5}\tau)\Big)d\tau.$

At this point things are getting messy for me because I don't know what I can do with the Dirac function right now and what property to use for it. If possible a hint to continue. Thank you

The value of $y=f(x)$ at $a$

Posted: 06 Mar 2022 07:28 AM PST

This is a strange question, based on a thought I had when dealing with integrals if given a variable $y$ such that $y=f(x)$ can we talk about the value of $y$ for a (separate) number $a$ if we were to substitute the value of $a$ for $x$ in the expression equal to $y$?

Case 1: Would this imply that $y=y_a=f(x)=f(a)$ as $x=a$ for e.g. for every value of $x$ it must be equal to $a$ or would it be a separate value $y_a=f(a)$ that can differ from $x$ (so it would be 'hypothetical') and could differ from $y$ itself.

Case 2: would it be that $y_a=f(a)$ is it's own independent value, and function of $a$ that would have the same value as $y$ when we want to assign the value of $a$ to $x$.

Perhaps asking what happens if we take the value of $a$ for $x$ imply that we are substituting $a$ for $x$ and hence receive an expression where $y$ = $f(a)$ for all or some values of $x, a$

Let $X \subset \Bbb R^2$ be a union of the coordinate axes and the line $x+y=1$, $0\le x \le1$. Show that $X$ is homotopy equivalent to $\Bbb S^1$.

Posted: 06 Mar 2022 07:38 AM PST

Let $X \subset \Bbb R^2$ be a union of the coordinate axes and the line $x+y=1$, $0\le x \le1$. Show that $X$ is homotopy equivalent to $\Bbb S^1$.

Denote the triangle formed by $(0,0),(1,0),(0,1)$ as $K$. The trick here is apparently to show that $K \simeq X$ and then using the fact that $K$ is homeomorphic to $\Bbb S^1$ to deduce that $X \simeq \Bbb S^1$.

With this I've managed to get the following. Define $f :X \to K$ as $$f(x) = \begin{cases} x, & x \in K \\ (0,1), &x \in \{0\} \times [1,\infty) \\ (1,0), & x \in [1,\infty) \times \{0\} \\ (0,0), & x\in \{0\} \times (-\infty, 0] \\ (0,0), &x \in (-\infty, 0] \times \{0\} \end{cases}$$ and define the inclusion $\iota :K \to X$. We know have that $f \circ \iota = id_K$ and I think I can define $h:K \times [0,1] \to X$ as $$h(a,t)=(1-t)(\iota \circ f)(a) + t \cdot id_X(a)$$ to show that $\iota \circ f \simeq id_X$?

The problem I'm having is that I didn't know that $K$ is homeomorphic to $\Bbb S^1$. What is the map giving this homeomorphism?

Every metric space is paracompact (an elegant proof)

Posted: 06 Mar 2022 07:42 AM PST

I'm studying a different proof to show that each metric space is paracompact in the book: Singh, Tej Bahadur-Introduction to Topology. It is a very elegant construction unlike the inductive method that seems more cumbersome to me. However there are three things that I could not understand:

(1) Why can some $E_{n, \alpha}$ be empty?

(2) How can I intuitively or geometrically see the sets $F_{n,\alpha}$ and $V_{n,\alpha}$?

(3) Why is it obtained that $F_{n,\alpha} \subset X - U_\beta $ or $F_{n,\beta} \subset X - U_\alpha$ in the last part of the proof?

I'm very interested in knowing a good argument for these questions since I have not been able to figure it out on my own after many attempts and I'm eager to know the answers. Any help is appreciated.

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Here are some definitions:

Definition. Let $X$ be a topological space. A collection of sets $\{U_{\alpha}\} \subset X$ (not necessarily open or closed) is said to be locally finite if to each ${x \in X}$, there is a neighborhood ${U}$ of ${x}$ that intersects only finitely many of the ${U_{\alpha}}$

Definition. Let ${\left\{U_{\alpha}\right\} }$ be a cover of a space ${X}$. Then a cover ${\left\{V_{\beta}\right\}}$ is called a refinement if each ${V_{\beta}}$ sits inside some ${U_{\alpha}}$

Definition. A Hausdorff space is paracompact if every open covering has a locally finite refinement.

Are these generalizations known in the literature?

Posted: 06 Mar 2022 07:45 AM PST

Using

$$\int_0^\infty\frac{\ln^{2n}(x)}{1+x^2}dx=\frac{\pi}{2^{2n+1}}\lim_{s\to \frac12}\frac{d^{2n}}{ds^{2n}}\csc(\pi s)=|E_{2n}|\left(\frac{\pi}{2}\right)^{2n+1}\tag{a}$$

where the $n$- derivative is explained here and

$$\text{Li}_{a}(-z)+(-1)^a\text{Li}_{a}\left(-\frac1z\right)=-2\sum_{n=0}^{\lfloor{a/2}\rfloor }\frac{\eta(a-2n)}{(2n)!}\ln^{2n}(z)\tag{b}$$

which follows from repeatedly integrating the common identity

$$\text{Li}_2(-z)+\text{Li}_2\left(-\frac1z\right)=-\frac{\ln^2(z)}{2}-2\eta(2)$$

after dividing by $z$, I managed to find:

$$\int_0^\infty\frac{\ln^{2n}(x)}{1+yx^2}dx=\frac{\left(\frac{\pi}{2}\right)^{2n+1}}{\sqrt{y}}\sum_{k=0}^n\binom{2n}{2k}|E_{2n-2k}|\pi^{-2k}\ln^{2k}(y)\tag{c}$$

$$\int_0^\infty\frac{\ln^{2n-1}(x)}{1+yx^2}dx=\frac{-\left(\frac{\pi}{2}\right)^{2n-1}}{2\sqrt{y}}\sum_{k=0}^{n-1}\binom{2n-1}{2k+1}|E_{2n-2k-2}|\pi^{-2k}\ln^{2k+1}(y)\tag{d}$$

$$\int_0^\infty\frac{\ln^{2n}(x)\text{Li}_{2a+1}(-x^2)}{1+x^2}dx=|E_{2n}|\left(\frac{\pi}{2}\right)^{2n+1}\zeta(2a+1)$$ $$-\frac{\left(\frac{\pi}{2}\right)^{2n+1}}{(2a)!}\sum_{k=0}^n \binom{2n}{2k}|E_{2n-2k}|\pi^{-2k}(2a+2k)!(2^{2k+2a+1}-1)\zeta(2k+2a+1)\tag{e}$$

$$\int_0^\infty\frac{\ln^{2n-1}(x)\text{Li}_{2a}(-x^2)}{1+x^2}dx=$$ $$-\frac{\left(\frac{\pi}{2}\right)^{2n-1}}{2(2a-1)!}\sum_{k=0}^{n-1} \binom{2n-1}{2k+1}|E_{2n-2k-2}|\pi^{-2k}(2a+2k)!(2^{2k+2a+1}-1)\zeta(2k+2a+1)\tag{f}$$

$$\sum_{k=1}^\infty\frac{(-1)^k H^{(2a+1)}_k}{(2k+1)^{2n+1}}=\frac{|E_{2n}|}{(2n)!}\left(\frac{\pi}{2}\right)^{2n+1}\zeta(2a+1)$$ $$-\frac{\left(\frac{\pi}{2}\right)^{2n+1}}{(2n)!(2a)!}\sum_{k=0}^n \binom{2n}{2k}|E_{2n-2k}|\pi^{-2k}(2a+2k)!(2^{2k+2a+1}-1)\zeta(2k+2a+1)\tag{g}$$

$$\sum_{k=1}^\infty\frac{(-1)^k H^{(2a)}_k}{(2k+1)^{2n}}=$$ $$\small{\frac{\left(\frac{\pi}{2}\right)^{2n-1}}{2(2a-1)!(2n-1)!}\sum_{k=0}^{n-1} \binom{2n-1}{2k+1}|E_{2n-2k-2}|\pi^{-2k}(2a+2k)!(2^{2k+2a+1}-1)\zeta(2k+2a+1)}\tag{h}$$

Question: Are the results of $(c)$ to $(h)$ known in the literature?

If the reader is curious about the correctness of the results above and wants to verify them on Mathematica, the Mathematica command of $|E_r|$ is Abs[EulerE[r]]

Thanks,

Straight line passes through a circle

Posted: 06 Mar 2022 07:31 AM PST

In a rectangular coordinate system, there is a circle $x^2 + y^2 -12 x+4 y-9=0$ and a straight line. If the straight line passes through $A(-2,-3)$ and intersects the circle at point $B$ and $C$, find the value of $A B \times A C$.

The given solution is this: $A B \times A C=\left(\sqrt{(-2)^{2}+(-3)^{2}-12(-2)+4(-3)-9}\right)^{2}=\left(\sqrt{16}\right)^{2}= \boxed{16}$

It looks like we simply substitute the values $(-2, -3)$ into the equation for the circle and find the square root of the result. How come? I don't quite understand.

How to differentiate inverse of a matrix inside frobenius norm?

Posted: 06 Mar 2022 07:34 AM PST

Basically, I have an equation in the form

$$ f = \left \| A^{-1} \right \|^2 $$

I need to differentiate the above equation wrt A.

The matrix A is just a 2X2 matrix so I tried to solve it by brute force. It worked but the solution is very long. Not to mention, it is very time-consuming. I was wondering if there is a general method to solve this problem that will work for any order of matrix A.

Volume of objects like hypercube / hypersphere : $V_{n}^{(m)}(r) = \dots$

Posted: 06 Mar 2022 07:48 AM PST

I am looking for some general form of equation for calculating volume for specific geometry objects.

The main idea is to find :

$$ V_{n}^{(m)}(r) = \dots $$ Where:
$V$ - volume of object
$n$ - regular polygon
$m$ - dimension
$r$ - radius of described sphere

It's easy to find equation for hypersphere, it is : $$ \lim_{n \to \infty} V_{n}^{(m)}(r) = \pi^{\frac{m}{2}} \frac{1}{\Gamma(\frac{m}{2} + 1)} * r^m $$ For $m=2$, general equation is : $$ V_{n}^{(2)}(r) = \frac{1}{2} n \sin(\frac{2 \pi}{n}) * r^2 $$

Thanks to Platonic solids there is equation for $m=3$ and $n=3$, $n=5$ : $$ V_{3}^{(3)}(r) = \frac{8 \sqrt{3}}{27} * r^3 $$ $$ V_{5}^{(3)}(r) = \frac{2 \sqrt{3} (5 + \sqrt{5})}{9} * r^3 $$

In general it's easy to see that equation will have form : $$ V_{n}^{(m)}(r) = f(n, m) * r^m $$

Is it possible to find exact equation ?

What worries me is limit in count of Platonic solids for $\text{3D}$ dimension case ($m=3$).
I wasn't also able to find any equation for other hyperobjects.

Positive integer solutions to product

Posted: 06 Mar 2022 07:28 AM PST

For $a,b,c,d\in\mathbb{N}$, I am looking for all positive integer solutions to $$a(b+1)c(d+1)=(a+1)b(c+1)d.$$ I already figured out that $a=b$ and $c=d$ as well as $a=d$ and $b=c$. But now I am stuck and I am wondering how I can check for more solutions. How can I approach this?

How to calculate start radius, end radius and angle for clothoid segments?

Posted: 06 Mar 2022 07:41 AM PST

I have a following construction:

Clothoid -- Circle Arc -- Clothoid -- Clothoid

which should form a smooth G2-continuous curve.

Clothoids are parameterized with clothoid parameter $A$ and length $L$. For the circle, its radius and arc length are given.

I should get the clothoids parameterized through start radius, end radius and angle. Can this be done?

I tried with usual formulas that can be found on wikipedia and on this site. For the first clothoid i took a big start radius i.e. $R=\infty$ and went on from there, taking the last end radius as the next starting one and so on. This didn't work as I thought.

List of pseudo-Catalan solids

Posted: 06 Mar 2022 07:20 AM PST

Convex solids can have all sorts of symmetries:

  • the platonic solids are vertex and face-transitive, meaning there is a subgroup of the rotations of 3-dimensional space which can bring any vertex onto another one (and the same for faces). The list there is limited to the 5 platonic solids.

  • face transitive (or isohedral) solids include the Catalan solids, the (infinite family) of dipyramids and the (infinite family) of trapezohedra. Note that without further restricitions these solids can come in infinite families: the rhombic dodecahedron has an infinite number of deltoidal cousins (see here ); it also fits in a one-parameter family of dodecahedra called pyritohedra (see here ); the dodecahedron and the triakis tetrahedron fit in the one-parameter family called tetartoid (see here ); dipyramids and trapezohedron also admit alls sorts of deformations beside the number of faces.

  • there is a much weaker symmetry one can ask for. Let's call it pseudo-Catalan (for lack of a better name). Fix a "centre" $C$. The convex solid is pseudo-Catalan, if each face can be sent to another face by a rotation with centre $C$ or a reflection (whose plane goes through $C$). Note that there is no requirement that this rotation (+ reflection) preserve the whole solid. An example of such a solid which is not a Catalan solid is the gyrate deltoidal icositetrahedron.

Question: is there a list of pseudo-Catalan solids?

  • note that there would be a last category, where the solid is convex and all the faces are congruent (a convex monohedral solid). The difference with the previous category is that translations are now allowed. In particular, to check that a solid belongs to the previous category, the choice of $C$ (and the fact that all rotations and reflections are constrained by this point) is important. Examples of such solids are the the triaugmented triangular prism and the gyroelongated square dipyramid.

[Edits: clarified definitions as to reflect the content of the comments below)]

How to differentiate a tensor Frobenius norm?

Posted: 06 Mar 2022 07:42 AM PST

I am wondering how to calculate

$$\nabla_{\mathcal{T}} \|\mathcal{T}-\mathcal{C}\|_F^2$$

where $\mathcal{T}, \mathcal{C} \in \mathbb{R}^{n\times n\times n\times n}$ are $4$th order tensors. Any help would be appreciated. Thank you.

Finding inverse probability density function.

Posted: 06 Mar 2022 07:25 AM PST

Let ${X \sim \!\, Exp(\lambda)}$, that is, ${X}$ is exponentially distributed with the probability density function ${f_X(x)= \lambda e^{-\lambda x}}, x≥0$. Determine the density and distribution functions for ${Y:=X^2}$

What i've done so far:

Exponentially distributed function by definition ${F_X(x)= 1- e^{-\lambda x}}$

Let ${g(x)= 1- e^{-\lambda x}}$, which is a continuous monotone strictly growing function.

Now I want to determine the density and distribution functions for ${Y:=g(X^2)=1- e^{-\lambda X}}$

${x^2=y \leftrightarrow x= \sqrt{y}}$ here is where I get stuck. Well, not sure if om on the right track. The answer is ${F_Y(y)= 1- e^{-\lambda \sqrt{y}}}$

And obviously the density function will be the derivative of the distribution function.

What needs to be done here?? thanks on beforehand.

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